arrays,file,audio,byte , What is the proper way to represent an audio sample in two bytes?

## Question:

Tag: arrays,file,audio,byte

I'm working on an application that generates a sequence of numbers representing a sine wave, and saves the sine wave in a `.wav` file to be played as audio.

The following line of code generates the value of a single sample in the audio wave:

``````double sample = Math.Sin(2.0 * Math.PI * frequency * i * (1.0 / samplesPerSecond));
``````

The resulting `.wav` file will be of 16 bit depth. Meaning, each sample is represented by 2 bytes (`short`). `double` however takes up 8 bytes.

What is the proper way to encode a `sample` into the resulting `byte[]`, where it can only occupy 2 bytes?

Since `sine` has a limited range `[-1.0, 1.0]`, you can map the value from that range to the range of 16-bit integers `[-32768, 32767]`.

``````short sample_short = SHORT_MAX * sample;
``````

There will be some round-off error, but that's necessary to squeeze the values into the less precise data type.

The formula is only this simple because the range of `sine` is from `-1` to `1`. If you were working with some other value that has a different range, you need to normalize it first:

``````result = DST_TYPE_MAX * original / ORIGINAL_MAX;
``````

This also assumes that the range of the original values is symmetric around `0`. If not, you need to do:

``````result = DST_TYPE_MAX * (original - ORIGINAL_MIN) / (ORIGINAL_MAX-ORIGINAL_MIN)
``````

This only works for values that have a limited range. You can't do this kind of normalization for a function like `tangent`, since its range is infinite.

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